Tüm alıştırma soruları

13931 soru

Soru 4021Soru

Simplify the trigonometric expression tan60+sin45cos45\frac{\tan 60^\circ + \sin 45^\circ}{\cos 45^\circ} to its exact surd form.

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Cevap: 6+1\sqrt{6} + 1

Cevap

6+1\sqrt{6} + 1
Substituting the exact values gives tan60=3\tan 60^\circ = \sqrt{3}, sin45=12\sin 45^\circ = \frac{1}{\sqrt{2}}, and cos45=12\cos 45^\circ = \frac{1}{\sqrt{2}}. Simplifying 3+1/21/2\frac{\sqrt{3} + 1/\sqrt{2}}{1/\sqrt{2}} gives 32+1=6+1\sqrt{3} \cdot \sqrt{2} + 1 = \sqrt{6} + 1.

Adım Adım Çözüm

1
Substitute the exact trigonometric values for the special angles
tan60=3\tan 60^\circ = \sqrt{3}, sin45=12\sin 45^\circ = \frac{1}{\sqrt{2}}, and cos45=12\cos 45^\circ = \frac{1}{\sqrt{2}}
Special angle values must be expressed in exact surd form.
2
Set up the fractional expression
3+1212\frac{\sqrt{3} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}
Replace each ratio with its exact surd equivalent.
3
Divide numerator terms by the denominator
312+1212=32+1=6+1\frac{\sqrt{3}}{\frac{1}{\sqrt{2}}} + \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{3} \cdot \sqrt{2} + 1 = \sqrt{6} + 1
Dividing by a fraction is equivalent to multiplying by its reciprocal.

Anahtar Kavram

Evaluation of Special Angle Trigonometric Ratios and Simplification of Surds
Soru 4022Soru

Consider the following list of numbers representing test scores of seven students: 1515, 44, 2222, 99, 1919, 22, and 1313. What is the median of this data set?

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Cevap: 1313

Cevap

The median of the data set is 1313.
First arrange the numbers in ascending order: 2,4,9,13,15,19,222, 4, 9, 13, 15, 19, 22. Since there are 77 numbers, the median is the middle (4th4\text{th}) number, which is 1313.

Adım Adım Çözüm

1
Arrange the given numbers in ascending order.
The ordered data set is 2,4,9,13,15,19,222, 4, 9, 13, 15, 19, 22.
To find the median of ungrouped data, the values must first be placed in numerical order.
2
Determine the position of the median for n=7n = 7 items.
Position = n+12=7+12=4th\frac{n + 1}{2} = \frac{7 + 1}{2} = 4\text{th} value.
For an odd number of observations nn, the median is the n+12th\frac{n+1}{2}\text{th} term.
3
Identify the 4th value from the ordered list.
The 4th value is 1313.
Counting from the smallest value, 22 is 1st, 44 is 2nd, 99 is 3rd, and 1313 is 4th.

Anahtar Kavram

Median of Ungrouped Data
Soru 4023Soru

A container holds a liquid mixture where water constitutes 38\frac{3}{8} of the total volume. When 15 liters15\text{ liters} of pure water is added to the mixture, water then accounts for 50%50\% of the new total volume. What was the initial total volume of the mixture in liters?

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Cevap: 60

Cevap

The initial total volume of the mixture was 60 liters.
The correct answer is 60 liters. By modeling the initial volume of water as 38V\frac{3}{8}V, adding 15 liters yields a new water volume of 38V+15\frac{3}{8}V + 15 out of a total volume of V+15V + 15. Setting 38V+15=0.5(V+15)\frac{3}{8}V + 15 = 0.5(V + 15) and solving gives V=60V = 60.

Adım Adım Çözüm

1
Define the variable and write an algebraic expression for the initial amount of water.
If VV is the initial total volume in liters, initial water volume = 38V\frac{3}{8}V.
Water makes up 38\frac{3}{8} of the total initial volume.
2
Account for the addition of 15 liters of pure water to both water volume and total volume.
New water volume = 38V+15\frac{3}{8}V + 15; New total volume = V+15V + 15.
Adding pure water increases both the specific water volume and the total mixture volume by 15 liters.
3
Formulate an equation relating new water volume to 50% of the new total volume.
38V+15=0.5(V+15)\frac{3}{8}V + 15 = 0.5(V + 15)
Water constitutes 50% (or 12\frac{1}{2}) of the updated mixture.
4
Solve the linear equation for VV.
7.5=18V    V=607.5 = \frac{1}{8}V \implies V = 60
Subtracting 38V\frac{3}{8}V and 7.57.5 from both sides isolates 18V\frac{1}{8}V on one side.

Anahtar Kavram

Solving multi-step fraction and percentage mixture problems
Soru 4024Soru

The table below shows the distribution of weights (in kg) of 10 packages in a warehouse:

Weight Interval (kg)Frequency (ff)
101410 - 1422
151915 - 1933
202420 - 2455

What is the mean weight of the packages?

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Cevap: 18.5 kg18.5\text{ kg}

Cevap

The mean weight of the packages is 18.5 kg18.5\text{ kg}.
The correct mean is calculated by finding the midpoint of each class interval (12,17,2212, 17, 22), multiplying each by its frequency to obtain fxfx (24,51,11024, 51, 110), and dividing the sum of fxfx (185185) by the total frequency (1010), giving 18.5 kg18.5\text{ kg}.

Adım Adım Çözüm

1
Calculate the class midpoint (xx) for each interval
Midpoints are: 10+142=12\frac{10+14}{2} = 12, 15+192=17\frac{15+19}{2} = 17, and 20+242=22\frac{20+24}{2} = 22.
For grouped data, each class interval is represented by its midpoint.
2
Multiply each midpoint (xx) by its corresponding frequency (ff) to find fxfx
2×12=242 \times 12 = 24, 3×17=513 \times 17 = 51, and 5×22=1105 \times 22 = 110.
This yields the total contribution of values from each class interval.
3
Sum the frequencies (f\sum f) and the products (fx\sum fx)
f=2+3+5=10\sum f = 2 + 3 + 5 = 10 and fx=24+51+110=185\sum fx = 24 + 51 + 110 = 185.
These sums are needed to compute the weighted mean.
4
Divide fx\sum fx by f\sum f
Mean xˉ=18510=18.5 kg\bar{x} = \frac{185}{10} = 18.5\text{ kg}.
The mean formula for grouped data is xˉ=fxf\bar{x} = \frac{\sum fx}{\sum f}.

Anahtar Kavram

Mean of Grouped Data using Class Marks
Soru 4025Soru

In a geometric progression of positive terms, the sum of the first two terms is 1212 and the sum of the third and fourth terms is 4848. What is the 6th6^{\text{th}} term of the progression?

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Cevap: 128

Cevap

The 6th term of the geometric progression is 128.
Dividing ar2(1+r)=48ar^2(1+r) = 48 by a(1+r)=12a(1+r) = 12 yields r2=4r^2 = 4, so r=2r = 2 for positive terms. Substituting r=2r = 2 into a(1+r)=12a(1+r) = 12 gives a=4a = 4. Using Tn=arn1T_n = a r^{n-1} for n=6n=6, we get T6=4×25=128T_6 = 4 \times 2^5 = 128.

Adım Adım Çözüm

1
Set up algebraic equations for the given sums using first term aa and common ratio rr.
a(1+r)=12a(1+r) = 12 and ar2(1+r)=48ar^2(1+r) = 48
The terms of a geometric progression are given by Tn=arn1T_n = a r^{n-1}.
2
Divide the equation for the third and fourth terms by the equation for the first and second terms.
r2=4    r=2r^2 = 4 \implies r = 2
Dividing eliminates aa and (1+r)(1+r), giving r2=4r^2 = 4. Since terms are positive, r>0r > 0.
3
Substitute r=2r = 2 into a(1+r)=12a(1+r) = 12 to solve for aa.
a=4a = 4
3a=123a = 12 leads directly to a=4a = 4.
4
Evaluate the 6th term using the formula T6=ar5T_6 = a r^{5}.
T6=4×25=128T_6 = 4 \times 2^5 = 128
Applying the general term formula Tn=arn1T_n = a r^{n-1} with n=6n=6.

Anahtar Kavram

Geometric Progression term relations and finding the common ratio from consecutive term pairs
Soru 4026Soru

In how many different ways can a chairperson and a secretary be selected from a committee of 55 members?

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Cevap: 2020

Cevap

The number of ways to select a chairperson and a secretary from 55 members is 2020.
Selecting 22 individuals for distinct positions (chairperson and secretary) from 55 candidates is an ordered selection problem. The number of ways is given by the permutation formula 5P2=5×4=20_5P_2 = 5 \times 4 = 20.

Adım Adım Çözüm

1
Identify whether order matters
Order matters because the roles of chairperson and secretary are distinct.
When distinct roles are assigned, the arrangement is a permutation rather than a combination.
2
Apply the permutation formula nPr=n!(nr)!_nP_r = \frac{n!}{(n-r)!} for n=5n=5 and r=2r=2
5P2=5!(52)!=5!3!=5×4=20_5P_2 = \frac{5!}{(5-2)!} = \frac{5!}{3!} = 5 \times 4 = 20
There are 55 choices for chairperson and 44 remaining choices for secretary.

Anahtar Kavram

Permutation of nn distinct items taken rr at a time
Tahmini Süre:45s
Soru 4027Soru

Find the set of real values of xx that satisfies both inequalities x23x10<0x^2 - 3x - 10 < 0 and 32x13 - 2x \le 1.

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Cevap: 1x<51 \le x < 5

Cevap

The set of real values satisfying both inequalities is 1x<51 \le x < 5.
Solving the quadratic inequality x23x10<0x^2 - 3x - 10 < 0 gives the open interval 2<x<5-2 < x < 5. Solving the linear inequality 32x13 - 2x \le 1 gives 2x2-2x \le -2, which upon dividing by 2-2 and reversing the inequality sign becomes x1x \ge 1. Finding the overlapping values that satisfy both inequalities gives 1x<51 \le x < 5.

Adım Adım Çözüm

1
Solve the quadratic inequality x23x10<0x^2 - 3x - 10 < 0.
Factor into (x5)(x+2)<0(x - 5)(x + 2) < 0. Critical values are x=2x = -2 and x=5x = 5. Since the inequality is strictly less than zero, the solution region is 2<x<5-2 < x < 5.
The quadratic expression is negative between its real roots.
2
Solve the linear inequality 32x13 - 2x \le 1.
Subtract 3 from both sides: 2x2-2x \le -2. Divide by 2-2 and flip the inequality sign: x1x \ge 1.
Dividing or multiplying an inequality by a negative number reverses the direction of the inequality sign.
3
Find the intersection of the two solution sets.
Combine 2<x<5-2 < x < 5 and x1x \ge 1 to get 1x<51 \le x < 5.
Values of xx must satisfy both conditions simultaneously.

Anahtar Kavram

Solving simultaneous linear and quadratic inequalities
Soru 4028Soru

What is the value of the expression 0.00048×0.0250.0016\frac{0.00048 \times 0.025}{0.0016} expressed in standard form?

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Cevap: 7.5×1037.5 \times 10^{-3}

Cevap

7.5×1037.5 \times 10^{-3}
The expression evaluates step-by-step to 0.75×1020.75 \times 10^{-2}, which when written in standard form A×10nA \times 10^n (where 1A<101 \le A < 10) gives 7.5×1037.5 \times 10^{-3}.

Adım Adım Çözüm

1
Convert each decimal in the expression to standard form (powers of 10).
0.00048=4.8×1040.00048 = 4.8 \times 10^{-4}, 0.025=2.5×1020.025 = 2.5 \times 10^{-2}, and 0.0016=1.6×1030.0016 = 1.6 \times 10^{-3}.
Converting decimals into standard form simplifies multiplication and division using laws of indices.
2
Simplify the numerator.
(4.8×104)×(2.5×102)=(4.8×2.5)×104+(2)=12×106=1.2×105(4.8 \times 10^{-4}) \times (2.5 \times 10^{-2}) = (4.8 \times 2.5) \times 10^{-4 + (-2)} = 12 \times 10^{-6} = 1.2 \times 10^{-5}.
Multiply the numerical coefficients together and add the exponents of 10.
3
Divide the numerator by the denominator.
1.2×1051.6×103=(1.21.6)×105(3)=0.75×102\frac{1.2 \times 10^{-5}}{1.6 \times 10^{-3}} = \left(\frac{1.2}{1.6}\right) \times 10^{-5 - (-3)} = 0.75 \times 10^{-2}.
Divide the coefficients and subtract the exponent of the denominator from that of the numerator.
4
Convert the result into standard form (A×10nA \times 10^n where 1A<101 \le A < 10).
0.75×102=(7.5×101)×102=7.5×1030.75 \times 10^{-2} = (7.5 \times 10^{-1}) \times 10^{-2} = 7.5 \times 10^{-3}.
Standard form requires the leading coefficient to be between 1 and 10.

Anahtar Kavram

Fractions, Decimals, and Standard Form Conversions
Soru 4029Soru

Given the function y=(x2+1)33x5y = \frac{(x^2 + 1)^3}{3x - 5}, calculate the value of dydx\frac{dy}{dx} at x=2x = 2.

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Cevap: -75

Cevap

The value of dydx\frac{dy}{dx} at x=2x = 2 is 75-75.
Applying the Quotient Rule uvuvv2\frac{u'v - uv'}{v^2} along with the Chain Rule to differentiate u(x)=(x2+1)3u(x) = (x^2+1)^3 yields u(x)=6x(x2+1)2u'(x) = 6x(x^2+1)^2. Evaluating at x=2x=2 gives u(2)=125u(2)=125, u(2)=300u'(2)=300, v(2)=1v(2)=1, and v(2)=3v'(2)=3, leading to 300(1)125(3)12=75\frac{300(1) - 125(3)}{1^2} = -75.

Adım Adım Çözüm

1
Set up the Quotient Rule framework
Let u(x)=(x2+1)3u(x) = (x^2 + 1)^3 and v(x)=3x5v(x) = 3x - 5, so that y=u(x)v(x)y = \frac{u(x)}{v(x)} and dydx=uvuvv2\frac{dy}{dx} = \frac{u'v - uv'}{v^2}.
The function is expressed as a quotient of two differentiable terms.
2
Differentiate the numerator using the Chain Rule
u(x)=3(x2+1)2ddx(x2+1)=6x(x2+1)2u'(x) = 3(x^2 + 1)^2 \cdot \frac{d}{dx}(x^2 + 1) = 6x(x^2 + 1)^2.
The numerator is a composite function requiring the inner derivative derivative of x2+1x^2+1 to be multiplied.
3
Differentiate the denominator
v(x)=3v'(x) = 3.
The derivative of a linear function 3x53x - 5 with respect to xx is its coefficient 3.
4
Evaluate u(x)u(x), u(x)u'(x), v(x)v(x), and v(x)v'(x) at x=2x = 2
u(2)=125u(2) = 125, u(2)=300u'(2) = 300, v(2)=1v(2) = 1, v(2)=3v'(2) = 3.
Substituting x=2x = 2 into each evaluated component simplifies the numerical calculation.
5
Substitute numerical values into the Quotient Rule formula
dydxx=2=(300)(1)(125)(3)12=3003751=75\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(3)}{1^2} = \frac{300 - 375}{1} = -75.
Completing the arithmetic calculation yields the final numerical derivative value.

Anahtar Kavram

Combining the Quotient Rule and Chain Rule for composite fractional functions
Tahmini Süre:1m 30s
Soru 4030Soru

The chairman criticized the committee's ephemeral interest in the reform policy, noting that their enthusiasm faded as soon as challenges arose. Which word is nearest in meaning to the underlined word as used in the sentence?

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Cevap: fleeting

Cevap

fleeting
The word 'ephemeral' describes something that lasts for a very short time. In the given sentence, the phrase 'their enthusiasm faded as soon as challenges arose' provides a clear contextual clue that the interest was brief. Therefore, 'fleeting' is the word nearest in meaning.

Adım Adım Çözüm

1
Analyze the context of the sentence
The sentence indicates that the committee's enthusiasm 'faded as soon as challenges arose', showing their interest was short-lived.
Contextual clues clarify the specific sense in which the target word is being used.
2
Determine the meaning of the underlined word 'ephemeral'
'Ephemeral' refers to something transitory, momentary, or lasting for a very short duration.
Matching the definition with contextual clues narrows down the exact synonym.
3
Evaluate the options against the target meaning
'Fleeting' directly conveys the meaning of lasting a very short time.
'Fleeting' is the accurate word nearest in meaning to 'ephemeral' in this scenario.

Anahtar Kavram

Contextual Synonyms in Lexis and Vocabulary
Soru 4031Soru

A trader buys a quantity of goods. He sells 14\frac{1}{4} of the total goods at a profit of 20%20\%, and 12\frac{1}{2} of the remaining goods at a loss of 10%10\%. If the rest of the goods are sold at cost price, resulting in an overall net profit of N1,500\text{N}1,500, what was the total cost price of the goods in Naira?

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Cevap: 120000

Cevap

The total cost price of the goods is 120,000 Naira.
Let CC be the total cost price of the goods. The first portion sold is 14C\frac{1}{4}C at a profit of 20%20\%, giving a gain of 0.20×14C=0.05C0.20 \times \frac{1}{4}C = 0.05C. The remaining portion is 114=34C1 - \frac{1}{4} = \frac{3}{4}C. Half of this remainder is 12×34C=38C\frac{1}{2} \times \frac{3}{4}C = \frac{3}{8}C, which is sold at a 10%10\% loss, causing a loss of 0.10×38C=0.0375C0.10 \times \frac{3}{8}C = 0.0375C. The final remaining 38C\frac{3}{8}C is sold at cost price (0 profit). Thus, the net profit is 0.05C0.0375C=0.0125C0.05C - 0.0375C = 0.0125C. Setting 0.0125C=1,5000.0125C = 1,500 and solving for CC gives C=1,5000.0125=120,000C = \frac{1,500}{0.0125} = 120,000 Naira.

Adım Adım Çözüm

1
Define the unknown total cost price
Let CC represent the total cost price of the goods in Naira.
Establishing a variable allows for algebraic formulation of profits and losses.
2
Calculate profit from the first portion
\text{Profit}_1 = 20\% \text{ of } \frac{1}{4}C = 0.20 \times 0.25C = +0.05C
The trader sells a quarter of the total value at a 20% gain.
3
Determine the remaining quantity and calculate loss from the second portion
\text{Remaining} = C - \frac{1}{4}C = \frac{3}{4}C; \quad \text{Second Portion} = \frac{1}{2} \times \frac{3}{4}C = \frac{3}{8}C = 0.375C; \quad \text{Loss}_2 = 10\% \text{ of } 0.375C = -0.0375C
The second sale applies to half of what was left after the first sale, incurred as a loss.
4
Equate net profit to given numerical value and solve for total cost price
\text{Net Profit} = 0.05C - 0.0375C = 0.0125C = 1,500 \implies C = \frac{1,500}{0.0125} = 120,000
The rest of the goods were sold at cost price (zero profit/loss), so net profit equals gain from portion 1 minus loss from portion 2.

Anahtar Kavram

Fractions and Percentages of Quantities
Tahmini Süre:1m 30s
Soru 4032Soru

A cubic curve is defined by the equation y=ax3+bx2+cx+dy = ax^3 + bx^2 + cx + d. The curve has a point of inflexion at (0,1)(0, 1) and a stationary point at (1,5)(1, 5). What is the local minimum value of the function?

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Cevap: 3-3

Cevap

The local minimum value of the function is 3-3.
By applying the conditions for a point of inflexion (y=0y''=0) and a stationary point (y=0y'=0), the curve equation is identified as y=2x3+6x+1y = -2x^3 + 6x + 1. Solving y=0y' = 0 gives x=1x = -1 for the local minimum, resulting in a minimum value of 3-3.

Adım Adım Çözüm

1
Use the point of inflexion and given point (0,1)(0, 1) to find bb and dd.
d=1d = 1 and b=0b = 0.
Since (0,1)(0,1) lies on the curve, substituting x=0,y=1x=0, y=1 gives d=1d = 1. The second derivative is y=6ax+2by'' = 6ax + 2b. At a point of inflexion x=0x = 0, y=0y'' = 0, giving 2b=0    b=02b = 0 \implies b = 0.
2
Use the stationary point condition at (1,5)(1, 5) to determine aa and cc.
a=2a = -2 and c=6c = 6.
The equation reduces to y=ax3+cx+1y = ax^3 + cx + 1. Since (1,5)(1,5) is on the curve, a+c+1=5    a+c=4a + c + 1 = 5 \implies a + c = 4. Also, y=3ax2+c=0y' = 3ax^2 + c = 0 at x=1x = 1, so 3a+c=03a + c = 0. Solving 3a+c=03a + c = 0 and a+c=4a + c = 4 simultaneously yields a=2a = -2 and c=6c = 6.
3
Find all stationary points of y=2x3+6x+1y = -2x^3 + 6x + 1 and test their nature using the second derivative.
Stationary points are at x=1x = 1 (maximum) and x=1x = -1 (minimum).
Setting y=6x2+6=0y' = -6x^2 + 6 = 0 yields x2=1    x=±1x^2 = 1 \implies x = \pm 1. The second derivative is y=12xy'' = -12x. For x=1x = -1, y=12>0y'' = 12 > 0, confirming a local minimum.
4
Evaluate the function at x=1x = -1 to find the local minimum value.
y(1)=3y(-1) = -3.
Substituting x=1x = -1 into y=2x3+6x+1y = -2x^3 + 6x + 1 yields y=2(1)3+6(1)+1=26+1=3y = -2(-1)^3 + 6(-1) + 1 = 2 - 6 + 1 = -3.

Anahtar Kavram

Determining curve constants using stationary points and points of inflexion, followed by identifying local extrema.
Tahmini Süre:3m 0s
Soru 4033Soru

A standard six-sided die was rolled 100100 times during a probability experiment, and the outcome 44 was recorded 2525 times. What is the experimental probability of rolling a 44?

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Cevap: 14\frac{1}{4}

Cevap

The experimental probability of rolling a 44 is 14\frac{1}{4}.
The experimental probability of an event is calculated by taking the ratio of the number of times the event occurs to the total number of trials performed. Since the outcome 44 appeared 2525 times out of 100100 rolls, the experimental probability is 25100\frac{25}{100}, which simplifies directly to 14\frac{1}{4}.

Adım Adım Çözüm

1
Identify the number of favorable trials and total trials
Favorable trials (rolling a 44) = 2525; Total trials = 100100.
Experimental probability depends on empirical data collected during the experiment.
2
Apply the experimental probability formula: P(E)=Frequency of EventTotal Number of TrialsP(E) = \frac{\text{Frequency of Event}}{\text{Total Number of Trials}}
P(rolling a 4)=25100P(\text{rolling a } 4) = \frac{25}{100}.
The experimental probability is defined as the relative frequency of the outcome.
3
Simplify the fraction to its lowest terms
25100=14\frac{25}{100} = \frac{1}{4}.
Dividing both the numerator and denominator by their greatest common divisor, 2525, gives the simplified fraction.

Anahtar Kavram

Experimental Probability
Soru 4034Soru

The second term of a geometric progression (G.P.) with positive terms is 66 and the fifth term is 4848. What is the sum of the first 66 terms of the progression?

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Cevap: 189189

Cevap

The sum of the first 66 terms is 189189.
Using the nn-th term formula Tn=arn1T_n = a r^{n-1}, we form two equations: ar=6a r = 6 and ar4=48a r^4 = 48. Dividing the fifth term equation by the second term equation gives r3=8r^3 = 8, so the common ratio r=2r = 2. Substituting r=2r = 2 back gives the first term a=3a = 3. Finally, applying the sum formula S6=3(261)21S_6 = \frac{3(2^6 - 1)}{2 - 1} gives 3×63=1893 \times 63 = 189.

Adım Adım Çözüm

1
Set up equations for the given terms using the nn-th term formula Tn=arn1T_n = a r^{n-1}.
T2=ar=6T_2 = a r = 6 and T5=ar4=48T_5 = a r^4 = 48.
The nn-th term of a G.P. is defined by Tn=arn1T_n = a r^{n-1}.
2
Divide the equation for T5T_5 by the equation for T2T_2 to find the common ratio rr.
\frac{a r^4}{a r} = \frac{48}{6} \implies r^3 = 8 \implies r = 2.
Dividing eliminates the first term aa and allows solving for rr directly.
3
Substitute r=2r = 2 into T2=6T_2 = 6 to find the first term aa.
a(2) = 6 \implies a = 3.
Knowing rr allows calculating aa from any known term.
4
Calculate the sum of the first 66 terms using Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}.
S_6 = \frac{3(2^6 - 1)}{2 - 1} = \frac{3(64 - 1)}{1} = 3 \times 63 = 189.
Applying the G.P. sum formula for n=6n = 6, a=3a = 3, and r=2r = 2 gives the total sum.

Anahtar Kavram

Geometric Progression term formula (Tn=arn1T_n = a r^{n-1}) and sum of nn terms formula (Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r - 1}).
Tahmini Süre:1m 30s
Soru 4035Soru

In a mathematics test, the mean score of a group of 1212 boys is 6060, and the mean score of a group of 1818 girls is 7070. What is the mean score of all 3030 students combined?

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Cevap: 66

Cevap

The combined mean score of all 30 students is 66.
To find the overall mean score for the entire class, determine the total sum of all test scores and divide by the total number of students. The total score contributed by the boys is 12×60=72012 \times 60 = 720, and the total score contributed by the girls is 18×70=126018 \times 70 = 1260. The combined total score is 720+1260=1980720 + 1260 = 1980. Dividing this total by 3030 students yields a combined mean of 6666.

Adım Adım Çözüm

1
Calculate the total score for the boys
Total boys' score = 12 × 60 = 720
The sum of data items equals the mean multiplied by the number of items.
2
Calculate the total score for the girls
Total girls' score = 18 × 70 = 1260
Multiply the number of girls by their mean score.
3
Find the combined total score and total student count
Combined total score = 720 + 1260 = 1980; Total students = 12 + 18 = 30
Sum the total scores and the total counts for both groups.
4
Calculate the combined mean score
Combined mean = 1980 / 30 = 66
Divide the combined total score by the total number of students.

Anahtar Kavram

Combined Mean of Two Ungrouped Datasets
Soru 4036Soru

Despite the intricate technical challenges and severe time constraints, the engineering team managed to __________ the complex structural renovation successfully. Which phrasal verb correctly completes the sentence?

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Cevap: bring off

Cevap

The phrasal verb 'bring off' correctly completes the sentence because it means to successfully perform or achieve something difficult.
The phrasal verb 'bring off' means to succeed in doing something difficult or unexpected. Given that the team faced technical challenges and tight deadlines, executing the renovation successfully is precisely expressed by 'bring off'.

Adım Adım Çözüm

1
Analyze the contextual clues in the sentence stem.
The sentence describes overcoming 'intricate technical challenges' and 'severe time constraints' to complete a renovation 'successfully'.
The blank requires a phrasal verb that signifies achieving or pulling off a challenging task.
2
Evaluate the particle meanings associated with the verb root 'bring'.
'Bring off' specifically denotes achieving success in an arduous undertaking, whereas 'bring about' implies causation, 'bring out' implies publication/emphasis, and 'bring up' implies mention/rearing.
Matching the precise idiomatic sense of the particle prevents misinterpretation of the verb's contextual role.

Anahtar Kavram

Distinguishing contextual meanings of phrasal verbs sharing the same root verb
Tahmini Süre:1m 0s
Soru 4037Soru

In a probability experiment, a card is drawn at random with replacement from a bag containing red, green, and blue cards. After conducting 250250 trials, a green card was drawn 8585 times. Given that the theoretical probability of drawing a green card is 0.300.30, calculate the positive difference between the experimental probability and the theoretical probability of drawing a green card.

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Cevap: 0.04

Cevap

The positive difference between the experimental probability and the theoretical probability is 0.04.
The experimental probability is calculated as the ratio of observed favorable outcomes to total trials: \(\frac{85}{250} = 0.34\). Subtracting the given theoretical probability of \(0.30\) yields a positive difference of \(|0.34 - 0.30| = 0.04\).

Adım Adım Çözüm

1
Determine experimental probability from trial data
Experimental probability = 85 / 250 = 0.34
Experimental probability is calculated as the ratio of observed favorable trials to the total number of trials executed.
2
Subtract theoretical probability from experimental probability
|0.34 - 0.30| = 0.04
Finding the positive difference requires subtracting the theoretical probability (0.30) from the experimental relative frequency (0.34).

Anahtar Kavram

Experimental probability is determined empirically by dividing the number of times an event occurs by the total number of trials, whereas theoretical probability is based on expected outcomes under ideal conditions.
Soru 4038Soru

A rectangular coil of 100100 turns with dimensions 0.10 m0.10\text{ m} by 0.20 m0.20\text{ m} is positioned perpendicular to a uniform magnetic field of 0.50 T0.50\text{ T}. The coil is rotated through 9090^\circ about an axis perpendicular to the field lines in a time interval of 0.040 s0.040\text{ s}, bringing its plane parallel to the magnetic field. What is the magnitude of the average electromotive force (e.m.f.) induced in the coil in volts?

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Cevap: 25

Cevap

The magnitude of the average induced electromotive force in the coil is 25 V25\text{ V}.
According to Faraday's law, the induced electromotive force magnitude is E=NΔΦΔtE = N \frac{\Delta \Phi}{\Delta t}. The initial flux through each turn is Φ1=BA=0.50 T×0.020 m2=0.010 Wb\Phi_1 = B A = 0.50\text{ T} \times 0.020\text{ m}^2 = 0.010\text{ Wb}. When rotated parallel to the field, the final flux Φ2\Phi_2 is 0 Wb0\text{ Wb}, so ΔΦ=0.010 Wb\Delta \Phi = 0.010\text{ Wb}. Substituting N=100N = 100 and Δt=0.040 s\Delta t = 0.040\text{ s} yields an induced e.m.f. of 100×0.0100.040=25 V100 \times \frac{0.010}{0.040} = 25\text{ V}.

Adım Adım Çözüm

1
Calculate the cross-sectional area of the rectangular coil.
A=0.10 m×0.20 m=0.020 m2A = 0.10\text{ m} \times 0.20\text{ m} = 0.020\text{ m}^2
The surface area is required to find the magnetic flux passing through each turn.
2
Determine the change in magnetic flux through one turn of the coil.
Initial flux Φ1=BA=0.50×0.020=0.010 Wb\Phi_1 = B A = 0.50 \times 0.020 = 0.010\text{ Wb}; Final flux Φ2=0 Wb\Phi_2 = 0\text{ Wb}; Change ΔΦ=0.010 Wb\Delta \Phi = 0.010\text{ Wb}
When perpendicular to the magnetic field, maximum flux links the coil. Rotating it parallel reduces the flux linking the coil to zero.
3
Apply Faraday's law of electromagnetic induction to calculate induced e.m.f.
E=NΔΦΔt=100×0.010 Wb0.040 s=25 VE = N \frac{\Delta \Phi}{\Delta t} = 100 \times \frac{0.010\text{ Wb}}{0.040\text{ s}} = 25\text{ V}
Faraday's law states that the induced e.m.f. magnitude equals the rate of change of total magnetic flux linkage.

Anahtar Kavram

Faraday's Law of Electromagnetic Induction
Soru 4039Soru

What are the values of yy for the real solution pairs (x,y)(x, y) that satisfy the simultaneous equations y3x=2y - 3x = 2 and y=x2x+5y = x^2 - x + 5?

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Cevap: 55 or 1111

Cevap

55 or 1111
Substituting y=3x+2y = 3x + 2 into the quadratic equation y=x2x+5y = x^2 - x + 5 gives 3x+2=x2x+53x + 2 = x^2 - x + 5, which simplifies to x24x+3=0x^2 - 4x + 3 = 0. Solving this yields x=1x = 1 and x=3x = 3. Substituting these into the linear equation gives y=3(1)+2=5y = 3(1) + 2 = 5 and y=3(3)+2=11y = 3(3) + 2 = 11. Therefore, the possible values of yy are 55 or 1111.

Adım Adım Çözüm

1
Express yy in terms of xx from the linear equation
y=3x+2y = 3x + 2
Linear equations can easily be substituted into quadratic equations.
2
Equate the linear expression for yy to the quadratic equation
3x+2=x2x+53x + 2 = x^2 - x + 5
Both expressions represent yy.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
x24x+3=0x^2 - 4x + 3 = 0
Subtract 3x3x and 22 from both sides.
4
Factorize and solve for xx
(x1)(x3)=0    x=1 or x=3(x - 1)(x - 3) = 0 \implies x = 1 \text{ or } x = 3
Finding the roots of the quadratic equation gives the xx-coordinates of the solution pairs.
5
Substitute each xx-value back into y=3x+2y = 3x + 2 to find yy
When x=1x = 1, y=3(1)+2=5y = 3(1) + 2 = 5. When x=3x = 3, y=3(3)+2=11y = 3(3) + 2 = 11.
The question specifically asks for the values of yy.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
Tahmini Süre:1m 30s
Soru 4040Soru

A liquid will boil when its saturated vapour pressure becomes equal to the prevailing external atmospheric pressure.

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Cevap: True

Cevap

The statement is True. A liquid boils when its saturated vapour pressure equals the external atmospheric pressure.
The statement accurately expresses the fundamental thermodynamic condition for boiling: the temperature of the liquid must reach a point where its saturated vapour pressure equals the surrounding atmospheric pressure.

Adım Adım Çözüm

1
Recall the definition of boiling point in thermal physics.
Boiling is the rapid conversion of liquid into gas occurring throughout the liquid body.
To determine the exact physical condition required for boiling to take place.
2
Relate saturated vapour pressure (SVP) to atmospheric pressure.
Bubbles of vapour can form within the liquid only when the pressure inside the bubbles (SVP) is equal to or greater than the pressure pushing down from the outside atmosphere.
If SVP is lower than atmospheric pressure, any vapour bubble attempting to form inside the liquid will immediately collapse.

Anahtar Kavram

Condition for Boiling of Liquids
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